If the probability distribution of a random variable $X$ is as follows,then $k=$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$2k$	$4k$	$3k$	$k$

(in $/10$)

In a city,it is found that $10$ accidents took place in a span of $50$ days. Assuming that the number of accidents follows the Poisson distribution,the probability that there will be $3$ or more accidents in a day in that city is

The variance of a Poisson variate $X$ is $2$. Then $P(X \geq 3) = $

If $m$ and $\sigma^2$ are the mean and variance of the random variable $X$,whose distribution is given by:

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

Then:

$A$ random variable $X$ has the following probability distribution:
| $x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|---|---|---|---|---|---|---|---|---|
| $P(x)$ | $0.15$ | $0.23$ | $0.12$ | $0.10$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the events $E = \{x \text{ is a prime number}\}$ and $F = \{x < 4\}$,the probability $P(E \cup F)$ is:

The probability distribution of a discrete random variable $X$ is given by the following table:

$X = x$	$0$	$1$	$2$	$3$	$4$
$P(X = x)$	$k$	$2k$	$4k$	$2k$	$k$

Then the value of $P(X \leq 2)$ is: